One way to see the topological nature of the system is through Laughlin charge pumping [1]}; on a two-dimensional cylinder with a perpendicular magnetic field, when a magnetic flux quantum is inserted through the cylinder, the state on the cylinder moves along the cylinder by one unit.
As we naturally have a cylindrical structure in the \(x\) -\(p_y\)  plane, it is natural to ask how Laughlin charge pumping appears in our setup.
As we see below, quantized pumping in our system occurs naturally due to construction.
A magnetic flux can be inserted through the cylinder in the \(x\) -\(p_y\)  plane by shifting the center of the harmonic trapping potential in the \(x\) -\(y\)  plane.
If we shift the central position of the harmonic trapping potentials by \(\delta x\) , the harmonic trapping potential acts as \(\frac{1}{2}m\omega ^2 (x-nx_0 -\delta x)^2\)  for a given tube labeled by an integer \(n\) . In the \(x\) -\(p_y\)  plane, this shift corresponds to the shift of the magnetic vector potential to \(\mathbf {A}(x,p_y) = (0, -x\tan \theta + \delta x \tan \theta )\) . The additional contribution of \(\delta x \tan \theta \)  is a constant term in the magnetic vector potential, and thus if we integrate it over the \(p_y\)  direction, we see that the cylinder in the \(x\) -\(p_y\)  plane is penetrated by a magnetic flux \(2\pi \delta x \tan \theta /y_0 = 2\pi \delta x / x_0\) . Therefore, inserting a magnetic flux quantum of \(2\pi \)  corresponds to taking \(\delta x = x_0\) . This shift of \(\delta x = x_0\)  corresponds to no change in the overall Hamiltonian except that a tube previously labeled by an integer \(n\)  now acts as a tube with a label \(n+1\) . Thus, as in the Laughlin pumping, the overall Hamiltonian comes back to the original one after the shift of \(\delta x = x_0\) . In the \(x\) -\(y\)  plane, the shift of \(\delta x = x_0\)  corresponds to moving the entire system by an amount \(\delta x = x_0\)  to the right, thus any state should also move horizontally by \(x_0\) . This implies that any state on the cylinder of the \(x\) -\(p_y\)  plane should also move along the cylinder by \(x_0\) , which is the manifestation of the Laughlin charge pumping in our setup.
Another way to see the topological nature of the system is through dynamical response of the system. In ultracold gases, quantized Hall response can be observed through the center-of-mass response upon application of a force (synthetic electric field) \(\mathbf {E} = (E_x, E_{p_y})\)  [2]}, [3]}, [4]}. When a synthetic electric field \(E_i\)  is applied along the \(i\)  direction, the center-of-mass velocity is [4]}
\(v^{\mathrm {C.M.}}_i = -\epsilon _{ij} \frac{2\pi }{A_{\mathrm {BZ}}}E_j \mathcal {C}\) ,
where \(A_{\mathrm {BZ}}\)  is the area of the Brillouin zone and \(\epsilon _{x p_y} = -\epsilon _{p_y x} = 1\) .
For our setup of a uniform magnetic field in the \(x\) -\(p_y\)  plane, there is no translational symmetry in the \(x\)  direction, and therefore we need to consider a magnetic unit cell and the magnetic Brillouin zone. A magnetic unit cell is an arbitrary rectangle in the \(x\) -\(p_y\)  plane which encloses one magnetic flux quantum. Since the strength of the artificial magnetic field is \(|B| = \tan \theta \) , the area of a magnetic unit cell is \(2\pi /\tan \theta \) . Then the area of the magnetic Brillouin zone is \((2\pi )^2\)  divided by the area of a magnetic unit cell and thus \(A_\mathrm {BZ} = 2\pi \tan \theta \) . Therefore, the center-of-mass velocity should follow
\(v^{\mathrm {C.M.}}_i = -\epsilon _{ij} E_j \mathcal {C}/\tan \theta = \epsilon _{ij} E_j \mathcal {C}/B. \) 
